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| 184_notes:i_b_force [2020/08/23 22:21] – dmcpadden | 184_notes:i_b_force [2021/06/16 21:58] – bartonmo | ||
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| {{youtube> | {{youtube> | ||
| - | ==== Force on a little chunk ==== | + | ===== Force on a little chunk ===== |
| If we think about a long straight wire with a //__steady state current__//, | If we think about a long straight wire with a //__steady state current__//, | ||
| Line 33: | Line 33: | ||
| Note: that the force is still given by the cross product between the $d\vec{l}$ and the $\vec{B}$, so the force on the piece of wire is //still// perpendicular to both the direction of the moving charges ($d\vec{l}$) and perpendicular to the magnetic field ($\vec{B}$). This means we can still use the [[184_notes: | Note: that the force is still given by the cross product between the $d\vec{l}$ and the $\vec{B}$, so the force on the piece of wire is //still// perpendicular to both the direction of the moving charges ($d\vec{l}$) and perpendicular to the magnetic field ($\vec{B}$). This means we can still use the [[184_notes: | ||
| - | ==== Force on the whole wire ==== | + | ===== Force on the whole wire ===== |
| Now that we have the magnetic force on a small piece of the wire, we can find the total force on the wire from the external magnetic field by adding up the contributions from each little piece of the wire. Since we have the small bit of force from the small bit of wire, we will add these using a integral: | Now that we have the magnetic force on a small piece of the wire, we can find the total force on the wire from the external magnetic field by adding up the contributions from each little piece of the wire. Since we have the small bit of force from the small bit of wire, we will add these using a integral: | ||
| $$\vec{F}_{wire}= \int_{wire} d\vec{F} = \int_{l_i}^{l_f} I d\vec{l} \times \vec{B}$$ | $$\vec{F}_{wire}= \int_{wire} d\vec{F} = \int_{l_i}^{l_f} I d\vec{l} \times \vec{B}$$ | ||